A mixture of aluminium and zinc weighing 1.67 grams was completely dissolved in acid and 1.69 litres of hydrogen gas was evolved. The hydrogen gas was measured at STP conditions. What was the mass percent of aluminium in original mixture?

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Solution

The correct option is D 75
Let x g of Al are present in sample. 1.67 - x gram of zinc will be present in sample. The atomic masses of Al and Zn are 27 g/mol and 65.4 g/mol respectively. The number of moles of Al present in sample $= \frac{x}{27}$ moles. The number of moles of Zn present in sample $= \frac{1.67 - x}{65.4}$ moles.
$2 A l + 6 H^{+} \to 2 A l^{3 +} + 3 H_{2}$
$Z n + 2 H^{+} \to Z n^{2 +} + H_{2}$
Volume of hydrogen obtained from Al $= \frac{3 \times 22.4 \times x}{2 \times 27}$ ......(1)
Volume of hydrogen obtained from Zn $= \frac{22.4 \times (1.67 - x)}{65.4}$ ......(2)
From (1) and (2)
$\frac{3 \times 22.4 \times x}{2 \times 27} + \frac{22.4 \times (1.67 - x)}{65.4} = 1.69$
On solving above equation, we get $x = 1.248$ g
Hence, the mass of Al in the sample is 1.248 g.
The percent of Al in the sample is $\frac{1.248}{1.67} \times 100 = 75$ %

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